Codeforces 588B Duff in Love【暴力】水题

B. Duff in Love

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Duff is in love with lovely numbers! A positive integer x is called lovely if and only if there is no such positive integera > 1 such that a² is a divisor ofx.

Malek has a number store! In his store, he has only divisors of positive integern (and he has all of them). As a birthday present, Malek wants to give her alovely number from his store. He wants this number to be as big as possible.

Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.

Input

The first and only line of input contains one integer, n (1 ≤ n ≤ 10¹²).

Output

Print the answer in one line.

Examples

Input

10

Output

10

Input

12

Output

6

Note

In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 islovely.

In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by4 = 2², so 12 is notlovely, while 6 is indeed lovely.

题目大意：

然你找到n的因子中（除了1以外）的最大的数，使得满足条件：

这个最大的因子数不能整除任意的平方数。

思路：

1、最大的因子可能是1e12.那么遍历平方数的时候需要O(sqrt(n))去遍历。

2、遍历找因子数的时候，同样需要O（sqrt(n)）去遍历。

如果两层放在一起就是O（n）了吗？

显然不是，我们知道，对于一个数来讲，其因子数的个数不会很多，所以暴力找因子数然后O（sqrt（n））去遍历平方数是不会超时的。

Ac代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>using namespace std;#define ll __int64int main(){    ll n;    while(~scanf("%I64d",&n))    {        ll output=0;        for(ll i=1;i<=sqrt(n);i++)        {            if(n%i==0)            {                int flag=0;                for(ll j=2;j<=1000000;j++)                {                    if(i%(j*j)==0)                    {                        flag=1;break;                    }                }                if(flag==0)output=max(output,i);                flag=0;                for(ll j=2;j<=1000000;j++)                {                    if((n/i)%(j*j)==0)                    {                        flag=1;break;                    }                }                if(flag==0)output=max(output,n/i);            }        }        printf("%I64d\n",output);    }}